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Imagine you're checking the temperature of water using a thermometer, the number that comes out in the form of degrees Celsius is a very good example of how we quantify concepts of physical quantities using special units. In the same way, when you measure the length of a table using a ruler, you are using concepts of units and measurement, which form the first chapter of the NCERT Solution Class 11 Physics Units and Measurement.
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Among students who are appearing in CBSE board exams, it is important that they master this chapter because there are a lot of questions that are directly based on NCERT concepts. It is due to this that study aids such as NCERT Solutions Class 11 Physics Chapter 1 can be very convenient. The Units and Measurement Class 11 NCERT solutions include clear and step-by-step explanations on each question of the exercises, as well as additional questions and formulas.
This Solution PDF contains the complete question and answer of Class 11 physics chapter 1, which will help during exam preparation, class tests, mock tests and homework-related work. Click the link below to download the PDF and use it offline according to your comfort.
Get the complete question and answer of this chapter, which contains fill-in-the-blanks, short answer, and long answer types of questions. Students are advised to solve the questions first by themselves, and if they are stuck at any step, then see the solution. These solutions are given with a step-by-step explanation.
Q1.1 Fill in the blanks
(a) The volume of a cube of side 1 cm is equal to ..... m3
(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ... (mm)2
(c) A vehicle moving with a speed of 18 kmh−1 covers....m in 1 s
(d) The relative density of lead is 11.3. Its density is .... gcm−3 or .... kgm−3 .
Answer:
(a) We know, 1cm=.01m (Tip: Divide by 100 to convert cm to m)
The volume of a cube of side a = a3 m3
Volume of cube of side 1 cm (i.e, .01 m ) = (.01)3=10−6 m3
(b) We know, 1cm=10mm (Tip: Multiply by 10 to convert cm to mm)
The surface area of a solid cylinder of radius r and height h = 2(πr2)+2πrh=2πr(r+h)
Required area = 2π(20)( 20+100)mm3=40π(120)mm3=4800π mm3=1.5×104 mm3
(c) (Tip: multiply by 5/18 to convert kmh−1 to ms−1 )
Distance covered = Speed×time=(18×5/18)ms−1×(1s)=5m
(d) Density = Relative Density × Density of water
(Density of water = 1 gcm−3=1000 kgm−3 )
Density of lead= (11.3)×1 gcm−3=11.3 gcm−3
Or, Density of lead=(11.3)×1000 kgm−3=11300 kgm−3=1.13×104 kgm−3
Q 1.2 : Fill in the blanks by suitable conversion of units
(a) 1kgm2s−2= gcm2s−2
b) 1m= ly
(c) 3.0 ms−2= kmh−2
(d) G=6.67×10−11Nm2(kg)−2= (cm)3s−2g−1
Answer:
(a) (1kg→1000g; 1m→100cm)
1kgm2s−2=(1000g)(100cm)2s−2=(103×104)gcm2s−2
⇒1kgm2s−2=107 gcm2s−2
(b) 1m=1.057×10−16ly (1 ly = Distance travelled by light in 1 year )
(c) (1m→10−3km; (60×60=)3600s→1hr)
3.0ms−2=3.0(10−3km)(1/3600hr)−2=3.0×(3600)2/1000 kmhr−2
⇒3.0ms−2=3.9×104 kmhr−2
(d) (N→kgms−2; 1m→100cm; 1kg→1000g)
G=6.67×10−11(kgms−2)m2kg−2=(6.67×10−11)(m)3s−2kg−1
⇒G=6.67×10−8cm3s−2g−1
Answer:
Given,
1Cal=4.2(kg)(m)2(s)−2
Given new unit of mass = αkg (In the old unit, 1kg corresponded to a unit mass, but in the new unit, αkg corresponds to a unit mass)
∴ In terms of the new unit, 1kg=1/α=α−1
Similarly, in terms of new units, 1m=1/β=β−1 and 1s=1/γ=γ−1
∴ 1Cal=4.2(kg)(m)2(s)−2=
1Cal=4.2(α−1)(β−1)2(γ−1)−2=4.2α−1β−2γ2
(a) atoms are very small objects
(b) a jet plane moves with great speed
(c) the mass of Jupiter is very large
(d) the air inside this room contains a large number of molecules
(e) a proton is much more massive than an electron
(f) the speed of sound is much smaller than the speed of light
Answer:
The given statement is true. A dimensional quantity may be small with respect to one reference and maybe large with respect to another reference. Hence, we require a standard reference to judge for comparison.
(a) An atom is a very small object with respect to a tennis ball. (but larger than an electron!)
(b) A jet plane moves with great speed with respect to a train.
(c) The mass of Jupiter is very large as compared to an apple.
(d) The air inside this room contains a large number of molecules as compared to in your lungs.
(e) A proton is much more massive than an electron
(f) The speed of sound is less than the speed of light
Answer:
Distance between Sun and Earth = Speed of light x Time taken by light to cover the distance
Speed of light = 1 unit
Time taken by light to reach Earth is 8 minute 20 seconds
Time taken = (8×60)+20=500s
The distance between Sun and Earth = 1 x 500 = 500 units.
Q 1.6 : Which of the following is the most precise device for measuring length :
(a) a vernier callipers with 20 divisions on the sliding scale
(b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale
(c) an optical instrument that can measure length to within a wavelength of light?
Answer:
To judge which tool is more precise, we have to find out their least count. Least count defines the margin of error and hence the precision. Hence the instrument with lower least count will be more precise.
(a) Least count = 1MSD−1VSD=1−(19/20)=1/20=0.05 cm (Taking 1 MSD as 1 mm)
(b) Least count = pitch/ number of divisions = 1/10000=0.001cm
(c) least count = wavelength of light = 400nm to 700nm, that is in the range of 10−7 m
Therefore, the optical instrument is the most precise device used to measure length.
Answer:
Given,
Magnification of Microscope = 100
The average width of hair under the microscope = 3.5 mm
(20 observations were made to calculate the average i.e. 3.5 mm as an experimental procedure. No need in our calculations.)
(Note: When magnified, the width is 3.5 mm. Hence actual width will be less by a factor of the magnification value.)
The average thickness of hair = 3.5mm100=0.035 mm.
Q 1.8 Answer the following :
(a)You are given a thread and a metre scale. How will you estimate the diameter of the thread?
(b)A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?
(c) The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only?
Answer:
(a) Take the thread and wrap it around the metre scale. Make sure the coils are packed closely without any space in between. If the diameter of the thread is d and number of turns obtained are n, then (n x d) corresponds to the marking on the metre scale, l.
Therefore, the diameter of the thread would be, d = l/n
(b) Theoretically, by increasing the number of divisions on the circular divisions, the value of least count decreases and hence accuracy increases.(Lower the value of least count, better will be the reading)
But practically, the number of divisions can be increased only up to a certain limit. (Also two adjacent divisions cannot be separated by a distance less than the human eye resolution!)
(c) With an increase in the number of observations, the accuracy of the experiment increases as the error is now distributed over a larger range. Hence, a set of 100 measurements of the diameter is expected to yield a more reliable estimate than a set of 5 measurements.
Answer:
Given,
Area of the house in the photo = 1.75 cm2
Area of the house on the screen = 1.55m2=1.55×104cm2
∴ Arial magnification, ma = Area on the screen / area on photo = 1.55×1041.75=0.886×104
Linear magnification of the projector-screen arrangement ml=ma=.886×104=94.1
Q 1.10 : State the number of significant figures in the following :
(a)0.007 m2
(b) 2.64×1024kg
(c) 0.2370 gcm−3
(d) 6.320 J
(e) 6.032 Nm−2
(f) 0.0006032 m2
Answer:
(a) The given value is 0.007 m2 .
Since, the number is less than 1 , the zeros on the right to the decimal before the first non-zero integer is insignificant. So, the number 7 is the only significant digit.
∴ It has 1 significant digit.
(b) The value is 2.64×1024kg
For the determination of significant values, we do not consider the power of 10 (Number is not less than 1). The digits 2, 6, and 4 are significant figures.
∴ It has 3 significant digits.
(c)The value is 0.2370 gcm−3 .
For the given value with decimals, all the numbers 2, 3, 7 and 0 are significant.
∴ It has 4 significant digits.
(d) The value is 6.320 J .
It has 4 significant digits.
(e) The value is 6.032 Nm−2 .
All the four digits are significant as the zeros in between two non-zero values are also significant.
∴ It has 4 significant digits.
(f) The value is 0.0006032 m2
Same as (a), first three zeroes after the decimal is insignificant. Only 6, 0, 3, 2 are significant.
∴ It has 4 significant digits.
Answer:
Given,
Length, l = 4.234 m ; Breadth, b = 1.005 m; Height, h = 2.01 cm = 0.0201 m
The length has 4 significant figures
The breadth has 4 significant figures
The height has 3 significant figures (Since the number is less than 1, hence zeroes after decimal before the first non-zero integer is insignificant)
We know,
Surface area of a cuboid, S = 2(l x b + b x h + h x l)
⇒ S= 2(4.234×1.005+1.005×0.0201+0.0201×4.234)
⇒ S= 2(4.255+0.020+0.085)
⇒ S= 8.72 m2
(Note: For addition/subtraction , the number of places after the decimal point in the answer is less than or equal to the number of decimal places in every term in the sum; i.e decided by the numbers after the decimal. )
Volume, V = l x b x h
⇒ V= 4.234×1.005×0.0201 = 0.0855 m3 ( ∴ 3 significant digits)
(Note: For Multiplication The LEAST number of significant digits in any number determines the number of significant figures in the answer; i.e decided by the number of significant digits )
The area has three significant values 2, 7 and 8.
The volume has three significant values 5, 5 and 8.
Answer:
Given,
The mass of the box = 2.30 kg
and the mass of the first gold piece = 20.15 g = 0.02015 kg
The mass of the second gold piece = 20.17 g = 0.02017 kg
(a) The total mass = 2.30 + 0.02015 + 0.02017 = 2.34032 kg
Since one is the least number of decimal places, the total mass = 2.3 kg.
(Note: For addition/subtraction , the number of places after the decimal point in the answer is less than or equal to the number of decimal places in every term in the sum; i.e decided by the numbers after the decimal. )
(b) Difference in masses =20.17−20.15=0.02 g
In subtraction, the final result should retain as many decimal places as there are in the number with the least decimal place
Guess where to put the missing c.
Answer:
The relation given is m=mo(1−v2)1/2
Divide both sides by mo ;
∴ L.H.S becomes m/mo which is dimensionless.
Hence, R.H.S must be dimensionless too. (After Dividing by mo !)
1(1−v2)1/2 can be dimensionless only when v→(v/c)
Therefore, the dimensional equation is m=mo(1−(vc)2)1/2
Q 1.14 : The unit of length convenient on the atomic scale is known as an angstrom and is denoted by: 1Å = 10^{-10} m$ . The size of a hydrogen atom is about 0.5Å. What is the total atomic volume in m3 of a mole of hydrogen atoms?
Answer:
Radius of an Hydrogen atom = 0.5 ÅÅ = 0.5 x 10 -10 m
Volume, V = 43πr3
⇒ V = 4/3×22/7×(0.5×10−10)3
⇒ V = 0.524×10−30m3
1 hydrogen mole contains 6.023×1023 hydrogen atoms.
The volume of 1 mole of hydrogen atom, V' = 6.023×1023×0.524×10−30
⇒ V' = 3.16×10−7m3≈3×10−7m3 .
Answer:
Radius of hydrogen atom = 0.5 Å = 0.5 x 10 -10 m (Size here refers to Diameter!)
Volume occupied by the hydrogen atom, V= 43πr3
⇒ V= 4/3×22/7×(0.5×10−10)3
⇒ V= 0.524×10−30m3
1 mole of hydrogen contains 6.023 x 10 23 hydrogen atoms.
Volume of 1 mole of hydrogen atom, V' = 6.023 x 10 23 x 0.524 x 10 -30
⇒ V' = 3.16 x 10-7 m3
Vm=22.4L=22.4×10−3m3
VmV′=22.4×10−33.16×10−7=7.09×104
The molar volume is 7.09×104 times greater than the atomic volume.
Hence, intermolecular separation in gas is much larger than the size of a molecule.
Answer:
Our eyes detect angular velocity, not absolute velocity. An object far away makes a lesser angle than an object which is close. That's why the moon (which is so far away!) does not seem to move at all angularly and thus seems to follow you while driving.
In other words, while in a moving train, or for that matter in any moving vehicle, a nearby object moves in the opposite direction while the distant object moves in the same direction. !
Answer:
Given,
Mass of the Sun, m = 3×1030 kg
The radius of the Sun, r = 8×108 m
∴ Volume V = 43πr3
⇒ V= 43×227×(8×108)3=2145.52×1024m3
Density = Mass/Volume
⇒ Density = 3×10302145.52×1024=1.39×103 kgm−3
Therefore, the density of the sun is in the range of solids and liquids and not gases. This high density arises due to inward gravitational attraction on outer layers due to inner layers of the Sun. (Imagine layers and layers of gases stacking up like a pile!)
Question 1:
The energy E and momentum p of a moving body of mass m are related by some equation. Given that c represents the need for light, identify the correct equation.
A. E2=pc2+m2c4
B. E2=pc2+m2c2
C. E2=p2c2+m2c2
D. E2=p2c2+m2c4
Answer:
Only quantities with the same dimension can be added or subtracted.
[E]=M1 L2 T−2[pc]=M1 L1 T−1⋅ L1 T−1=M1 L2 T−2[mc2]=M1 L2 T−2The only correct option is DE2=p2c2+m2c4
Question 2:
A physical quantity Q is related to four observables a,b,c,d as follows : Q=ab4cd where, a=(60±3)Pa;b=(20±0.1)m; c=(40±0.2)Nsm−2 and d=(50±0.1)m, then the percentage error in Q is x1000, where x= _____ .
Answer:
Q=ab4 cd
⇒ΔQQ×100=[Δaa+4Δ b b+Δcc+Δdd]×100
⇒x1000=[360+4(0.120)+(0.240)+0.150]×100
⇒x=7700
Question 3:
The least count of a screw gauge is 0.01 mm. If the pitch is increased by 75% and the number of divisions on the circular scale is reduced by 50%, the new least count will be _________ ×10−3 mm.
Answer:
Given least count of Screw Gauge =0.01 mm
L. C=( pitch ) No. of circular turn =P N=0.01 mm
New pitch, p' =P(1+0.75)N(1−0.5)=P N[1.750.5]
⇒p′=(0.01)3.5
⇒p′=0.035 mm
⇒p′=35×10−3 mm
Question 4:
A tiny metallic rectangular sheet has a length and breadth of 5 mm and 2.5 mm, respectively. Using a specially designed screw gauge which has a pitch of 0.75 mm and 15 divisions in the circular scale, you are asked to find the area of the sheet. In this measurement, the maximum fractional error will be x100 where x is _________.
Answer:
Since the least count of the instrument can be calculated as
Least count, LC = pitch length No. of division on a circular scale
⇒LC=0.7515=0.05 mm
Here we are provided L=5 mm&W=2.5 mm
∵ We know that
A=L.W
For the calculation of fractional error, we can write
dA A=dL L+d W W
dA A=0.055+0.052.5⇒dA A=1100+2100=3100
So, x=3
Question 5:
The maximum percentage error in the measurement of the density of a wire is
[Given, mass of wire =(0.60±0.003)g, radius of wire =(0.50±0.01)cm, length of wire (10.00±0.05)cm]
Answer:
Given,
mass of wire =(0.60±0.003)g
radius of wire =(0.50±0.01)cm
length of wire (10.00±0.05)cm
We know,
ρ=mV=mπr2l
Δρρ%=Δmm×100+2(Δrr)×100+Δll×100=0.0030.60×100+2(0.010.50)×100+0.0510.00×100=0.5+4+0.5=5%
Chapter 1 Units and Measurements introduces you to the most fundamental units of physical science, that is, the defining, measuring, and expressing physical quantities.
1.1 Introduction
1.2 The International System Of Units
1.3 Significant Figures
1.3.1 Rules For Arithmetic Operations With Significant Figures
1.3.2 Rounding Off The Uncertain Digits
1.3.3 Rules For Determining The Uncertainty In The Results Of Arithmetic Calculations
1.4 Dimensions Of Physical Quantities
1.5 Dimensional Formulae And Dimensional Equations
1.6 Dimensional Analysis And Its Applications
1.6.1 Checking The Dimensional Consistency Of Equations
1.6.2 Deducing the Relation Among The Physical Quantities
The significant formulae of this chapter enable the students to effectively approach numerical exercises, verify the accuracy of physical relations and convert units appropriately. Good mastery of such formulae is important in making a foundation in physics.
Least Count of an instrument = Smallest division on the scale
Example:
Absolute error =Σ(|ai−amean |)n Relative error =Δamaan amax Percentage error =Δamaan amax ×100
In order to effectively answer questions in the Units and Measurement chapter, one should first of all make sure that he/she has a clear understanding of the basic quantities in physics and their standard units, specifically in the SI system. Apply the appropriate formulas and make sure that the quantities of units match. None of the calculations should be made before all of the quantities are in proper units. When dealing with significant numbers, pay great attention to them since they show the accuracy of measurements. In the case of errors, use absolute, relative or percentage errors as per the case requirement. In questions where dimensional analysis is applicable, undertake the method of dimensional analysis in order to test the validity of equations used or to obtain relations between physical quantities.
Below is a comparison table of the concepts given, which are not in the NCERT syllabus but present in the JEE Main syllabus. Students who are targeting the JEE Main should go through the extra concepts which are mentioned in the table.
Apart from the chapterwise class 11 physics solution, students can also access the NCERT book and syllabus for class 11 by clicking on the article listed below.
It creates the basis of comprehension of physical quantities and provides precise scientific communication.
Basic physical quantities (such as meter, kilogram) are fundamental units and derived units can be created using fundamental units (such as m/s which is a unit of velocity).
Yes, it is basic and is often asked during examinations such as NEET, JEE and school-level Olympiad.
Accuracy is the degree to which a measurement approaches the true value, whereas precision is the degree to which repeated measurements are close to each other.
You can deduce relations and ensure equations are consistent by comparing the dimensions of physical quantities.
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